# finite projective planes have $q^{2}+q+1$ points and $q^{2}+q+1$ lines

Given a finite projective plane that contains a quadrangle OXYZ (i.e. no three of these four points are on one line). To prove: the plane has $q^{2}+q+1$ points and $q^{2}+q+1$ lines for some integer $q$, and there are $q+1$ points on each line and $q+1$ lines through each point.

Let $x$ and $y$ be the lines OX and OY, which must exist by the axioms. By the assumption  OXYZ is a quadrangle these lines are distinct and Z is not on them. Let there be $p$ points X${}_{i}$ on $x$ other than O, for each of them one line ZX${}_{i}$ exists, and is distinct (one lines cannot pass through two X${}_{i}$ unless it is $x$ but that’s not a line through Z). Conversely every line through Z must intersect $x$ in a unique point (two lines intersecting in Z cannot intersect at another point, and Z is not a point on $x$). So there are $p+1$ lines through Z (OZ is one of them). By the same reasoning, using $y$, there are $q+1$ lines through Z so $p=q$. We also found $q+1$ points (including O) on $y$ and the same number on $x$. Intersecting the $q+1$ lines through Z with XY (on which Z does not lie, the quadrangle again) reveals at least $q+1$ distinct points there and at most $q+1$ because for each point there there is a line through it and Z.

The lines not through O intersect $x$ in one of the $q$ points X${}_{i}$ and $y$ in one of the $q$ points Y${}_{j}$. There are $q^{2}$ possibilities and each of them is a distinct line, because there is only one line through a given X${}_{i}$ and Y${}_{j}$. The lines that do pass through O intersect XY in one of the $q+1$ points there, again one line for each such point and vice versa. That’s $q+1$ lines through O and $q^{2}$ not through O, $q^{2}+q+1$ in all.

There are $q+1$ lines through X (to each of the points of $y$) and $q+1$ lines through Y (to each of the points of $x$). Intersect the $q$ lines through X other than XY with the $q$ lines through Y other than XY, these $q^{2}$ intersections are all distinct because for any P there’s only one line PX and one line PY. Note we did not use the line XY. Conversely for any P not on XY there must be some PX and some PY, so there are exactly $q^{2}$ points not on XY. Add the $q+1$ points on XY for a total of $q^{2}+q+1$.

The constructions above already showed $q+1$ lines through some points (X, Y and Z), by the same games as before that implies for each of them $q+1$ points on every line not through that point. We also saw $q+1$ points on some lines ($x$, $y$, XY) which implies for each of them $q+1$ lines through every point not on that line. Such reasoning covers $q^{2}$ items on first application and rapidly mops up stragglers on repeated application.

Some form of this proof is standard math lore; this version was half remembered and half reconstructed.

Title finite projective planes have $q^{2}+q+1$ points and $q^{2}+q+1$ lines FiniteProjectivePlanesHaveQ2q1PointsAndQ2q1Lines 2013-03-22 15:11:18 2013-03-22 15:11:18 marijke (8873) marijke (8873) 4 marijke (8873) Proof msc 51A35 msc 05B25 msc 51E15